unit 4 : exit intermediate 2 – additional question bank social arithmetic logic diagrams formulae

UNIT 4 :

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INTERMEDIATE 2 – ADDITIONAL QUESTION BANK

SocialArithmetic

Logic Diagrams

Formulae

UNIT 4 :

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Please choose a question to attempt from the following:

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SocialArithmetic

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2 3 4

SOCIAL ARITHMETIC: Question 1

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Joanne is a sales assistant who earns a basic wage of £180 per week plus 10% commission on all sales over £200.

(a) How much does she earn in a week when her sales are £530?

(b) What do her sales need to be so her wages are £250?


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(b) What do her sales need to be so her wages are £250?

1. Calculate how much of the sales qualify for a commission

payment.

2. Apply given percentage to this

sum to find Commission due.

3. Remember total Wage includes

basic

4. In (b) use reverse percentage to work

out sales required to give the amount above

basic.

5. In (b) Remember sales must be £200 before commission earned.

What would you like to do now?


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(b) What do her sales need to be so her wages are £250? = £330

= £900


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Question 1

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Joanne is a sales assistant

who earns a basic wage of

£180 per week plus 10%

commission on all sales

over £200.


1. Calculate how much of the sales qualify for a commission payment.

(a)Commissionable sales

= £530 - £200

= £330

2. Apply given percentage to this sum to find Commission due.

Commission = 10% of £330

= £33

3. Remember total Wage includes basic.Continue Solution

Total wage = £180 + £33

= £213

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Question 1

Soc Arith Menu

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Joanne is a sales assistant

who earns a basic wage of

£180 per week plus 10%

commission on all sales

over £200.


2. Use reverse percentage to work out sales required to give this figure.

3. Remember sales must be £200 before commission earned.Continue Solution

(b) What do her sales need to

be so her wages are £250?

(b)Total commission

= £250 - £180

= £70

Sales on which commission earned

= 10 x £70 = £700

Total sales = £700 + £200

= £900


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Percentage calculations:

10% of £330 = x 330

= 0.10 x 330

etc.

10100


(a)Commissionable sales

= £530 - £200

= £330

2. Apply given percentage to this sum to find Commission due.


= £33

3. Remember total Wage includes basic.

Total wage = £180 + £33

= £213

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X 10 X 10

Working backwards (reverse %)

From 10% require 100%


2. Use reverse percentage to work out sales required to give this figure.

3. Remember sales must be £200 before commission earned.

(b)Total commission

= £250 - £180

= £70

Sales on which commission earned

= 10 x £70 = £700

Total sales = £700 + £200

= £900

10% = £70

100% = £700

(b) Kerry’s monthly pension contributions are 6.5% of her gross

salary. Find this and hence find her net salary for May.

Kerry Owen works for a builders’ supplies merchant. Her partly completed payslip for May is shown below

(a) Kerry’s basic monthly salary is £1700 plus overtime plus commission of 2% on all her sales. Find her gross salary for May when her sales totalled £43600.

Name Employee No. N.I. No. Tax Code Month

Basic salary Commission Overtime Gross salary

Nat.Insurance Income tax Pension Total Deductions

Net salary

K.Owen 34/09852 KU34521D 498H May

£1700 £240.58

£185.63 £487.25


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(b) Kerry’s monthly pension contributions are 6.5% of her gross

salary. Find this and hence find her net salary for May.

Kerry Owen works for a builders’ supplies merchant. Her partly completed payslip for May is shown below

(a) Kerry’s basic monthly salary is £1700 plus overtime plus commission of 2% on all her sales. Find her gross salary for May when her sales totalled £43600.

Name Employee No. N.I. No. Tax Code Month

Basic salary Commission Overtime Gross salary


Net salary

K.Owen 34/09852 KU34521D 498H May

£1700 £240.58

£185.63 £487.25


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EXIT

= £2812.58

= £182.82

= £1956.88What would you like to do now?

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Question 2

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Kerry’s basic monthly salary

is £1700 plus overtime plus

commission of 2% on all her

sales. Find her gross salary

for May when her

sales totalled £43600.

(a)


= 0.02 x £43600

= £872

Gross salary = £1700 + £872 + £240.58

= £2812.58Back to payslip

Overtime!

(b)

Kerry’s monthly pension contributions are 6.5% of her gross salary.

Find this and hence find her net salary for May.

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Question 2

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(b)

Pension = 6.5% of £2812.58

= 0.065 x £2812.58

= £182.82 to nearest penny

Found in part (a)

Total deductions

= £185.63 + £487.25 + £182.82

= £855.70

Net salary = £2812.58 - £855.70

= £1956.88

(see payslip)


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Percentage Calculations

2% of £43600 = x £43600

= 0.02 x £43600

2100

(a)


= 0.02 x £43600

= £872

Gross salary = £1700 + £872 + £240.58

= £2812.58

Overtime!Gross Pay =

Basic + Commission + Overtime

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(b)

Pension = 6.5% of £2812.58

= 0.065 x £2812.58

= £182.82 to nearest penny

Total deductions

= £185.63 + £487.25 + £182.82

= £855.70

Net salary = £2812.58 - £855.70

= £1956.88

(see payslip)

Net pay =

Gross Pay

- (Nat. Ins. + Inc Tax + Pension)

TOTAL DEDUCTIONS


Name Employee No. N.I. No. Tax Code Week

Basic wage Bonus Overtime Gross wage


Net wage

D.Marr 2001/0789 WA12311F 395L 37

£265.65

£26.32 £60.93 £45.83

£41.35 £58.70


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Diana Marr works in an electronics factory. Her partially completed payslip is shown below.

(a) Find her gross wage for this particular week.

(b) If she works a 38 hour basic week then find her hourly rate.

Name Employee No. N.I. No. Tax Code Week

Basic wage Bonus Overtime Gross wage


Net wage

D.Marr 2001/0789 WA12311F 395L 37

£265.65

£26.32 £60.93 £45.83

£41.35 £58.70


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EXIT

Diana Marr works in an electronics factory. Her partially completed payslip is shown below.

(a) Find her gross wage for this particular week.

(b) If she works a 38 hour basic week then find her hourly rate.


= £398.73

= £7.86

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Question 3

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(a)

Find her gross wage for

this particular week.

Back to payslip

(a)Total deductions

= £26.32 + £60.93 + £45.83

= £133.08

Gross wage = £133.08 + £265.65

= £398.73

Work backwards!

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Question 3

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(b)

If she works a 38 hour

basic week then find her

hourly rate.

(b)Hourly rate = Basic wage

Hours worked

Basic wage

= £398.73 - £58.70 - £41.35

= £298.68

Hourly rate = £298.68 38

= £7.86

Gross from part (a)

Basic wage is before overtime and bonus.


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Working Backwards

Gross pay

= Net pay + Deductions

(a)Total deductions

= £26.32 + £60.93 + £45.83

= £133.08

Gross wage = £133.08 + £265.65

= £398.73

Work backwards!

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Basic wage

= £398.73 - £58.70 - £41.35

= £298.68

Hourly rate = £298.68 38

= £7.86

Gross from part (a)

Basic wage is before overtime and bonus. Working Backwards

Hourly rate = Basic wage

Hours worked


Amount 60 months 48 months 24 months

LP Basic LP Basic LP Basic

£8000 169.83 166.51 204.04 200.03 376.23 368.86

£6000 127.38 124.88 153.03 150.03 282.18 276.66

£4000 84.92 83.26 102.02 100.02 188.12 184.43

£2000 42.46 41.63 51.01 50.01 94.06 92.22

SOCIAL ARITHMETIC: Question 4 EXIT

A couple are having new windows fitted. The following table shows the monthly repayment charges on various amounts.

(a) They borrow £6000 over 4 years and decide to make basic repayments. How much do they actually pay back?

(b) How much extra would they repay if they had opted for a 5 year repayment period with loan protection?

LP – Loan Protection

Amount 60 months 48 months 24 months


£8000 169.83 166.51 204.04 200.03 376.23 368.86

£6000 127.38 124.88 153.03 150.03 282.18 276.66

£4000 84.92 83.26 102.02 100.02 188.12 184.43

£2000 42.46 41.63 51.01 50.01 94.06 92.22

SOCIAL ARITHMETIC: Question 4 EXIT

A couple are having new windows fitted. The following table shows the monthly repayment charges on various amounts.



LP – Loan Protection

= £7201.44

= £441.36

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Question 4

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(a) They borrow £6000 over

4 years and decide to make

basic repayments.

How much do they actually

pay back?

Back to table

(a)

Monthly repayment = £150.03

Total repayment = £150.03 x 48

= £7201.44

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Question 4

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(b)

How much extra would they

repay if they had opted for

a 5 year repayment period

with loan protection?

= £7201.44Answer from (a)

= £7642.80

= £441.36

(b) Monthly repayment = £127.38


Extra paid = £7642.80 - £7201.44


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(a)



= £7201.44

Be careful when using tables that you identify relevant

categories. In this case:1. Amount

2. Repayment period3. Loan Protection

Additional Comments

AMOUNT BORROWED

60 MONTHS 48 MONTHS 24 MONTHS


£8000 169.83 166.51 204.04 200.03 376.23 368.86

£6000 127.38 124.88 153.03 150.03 282.18

276.66 £4000 84.92 83.26 102.02 100.02 188.12 184.43 £2000 42.46 41.63 51.01 50.01 94.06 92.22

LP - Loan Protection

4years = 4 x 12 = 48 months

So required monthly repayment = £150.03


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(a)



= £7201.44

Total Repayment =

monthly instalment

x no. of instalments

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= £7642.80

= £441.36



Extra paid = £7642.80 - £7201.44

Be careful when using tables that you identify relevant

categories. In this case:1. Amount

2. Repayment period3. Loan Protection

AMOUNT BORROWED

60 MONTHS 48 MONTHS 24 MONTHS


£8000 169.83 166.51 204.04 200.03 376.23 368.86

£6000 127.38 124.88 153.03 150.03 282.18

276.66 £4000 84.92 83.26 102.02 100.02 188.12 184.43 £2000 42.46 41.63 51.01 50.01 94.06 92.22

LP - Loan Protection

5 years = 60 months



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= £7642.80

= £441.36



Extra paid = £7642.80 - £7201.44

Extra Repaid =

Cost under option 1

- Cost under option 2

UNIT 4 :



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Formulae

1

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2 3 4 5

FORMULAE: Question 1

The surface area of a triangular prism, S cm2, is given by the formula

S = x2 + 2dx + dw

where all distances are in cm.

x x

d

w

(a) Find S when x = 10, w = 14 & d = 30.

(b) Find w when x = 5, d = 20 & S = 365.

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The surface area of a triangular prism, S cm2, is given by the formula

S = x2 + 2dx + dw

where all distances are in cm.

x x

d

w

(a) Find S when x = 10, w = 14 & d = 30.

(b) Find w when x = 5, d = 20 & S = 365.

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w = 7

= 1120 cm2

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Question 1

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(a)

S = x2 + 2dx + dw

= (10 x 10) + (2 x 30 x 10) + (30 x 14)

= 100 + 600 + 420

= 1120

Area is 1120cm2

1. Substitute known values into given formula:

S = x2 + 2dx + dw

(a)Find S when x = 10,

w = 14 & d = 30.

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Question 1

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S = x2 + 2dx + dw

(b)Find w when x = 5,

d = 20 & S = 365.

(b)

S = x2 + 2dx + dw

365 = (5 x 5) + (2 x 20 x 5) + (20 x w)

20w + 200 + 25 = 365

20w = 140

w = 7

Width is 7cm

2. Tidy up then solve equation for target letter:


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1. Write formula

2. Replace known values

3. Evaluate

(a)

S = x2 + 2dx + dw

= (10 x 10) + (2 x 30 x 10) + (30 x 14)

= 100 + 600 + 420

= 1120

Area is 1120cm2

When evaluating formulae:

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(b)

S = x2 + 2dx + dw

365 = (5 x 5) + (2 x 20 x 5) + (20 x w)

20w + 200 + 25 = 365

20w = 140

w = 7

Width is 7cm

2. Tidy up then solve equation for target letter:

1. Write formula


3. Evaluate


4. Solve resulting equation



EXIT

Full solution

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Answer

To convert temperatures from °F into °C we use the formula

C = 5/9(F - 32)

(a) Change 302°F into °C .

(b) Change -40°F into °C , and comment on your answer.

(c) Change 10°C into °F .


EXIT

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To convert temperatures from °F into °C we use the formula

C = 5/9(F - 32)

(a) Change 302°F into °C .

(b) Change -40°F into °C , and comment on your answer.


302°F = 150°C

-40°F = -40°C

10°C = 50°F


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Question 2

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(a) C = 5/9(F - 32)

C = 5/9(302 - 32)

C = 5/9 of 270

C = 270 9 x 5 = 150

302°F = 150°C

BODMASC = 5/9(F - 32)

(a) Change 302°F into °C

(b) Change -40°F into °C


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Question 2

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BODMASC = 5/9(F - 32)




(b) C = 5/9(F - 32)

C = 5/9(-40 - 32)

C = 5/9 of -72

C = -72 9 x 5 = -40

-40°F = -40°C

Value same in each scale!!

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Question 2

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C = 5/9(F - 32)




(c) C = 5/9(F - 32)

10 = 5/9(F - 32) (x9)

90 = 5(F - 32)

90 = 5F - 160

5F = 90 + 160

5F = 250

F = 50

10°C = 50°F


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(a) C = 5/9(F - 32)

C = 5/9(302 - 32)

C = 5/9 of 270

C = 270 9 x 5 = 150

302°F = 150°C

BODMAS

1. Write formula


3. Evaluate ( BODMAS)


BO÷ / x+ / -

BracketsOf÷ / x+ / -

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BODMAS(b) C = 5/9(F - 32)

C = 5/9(-40 - 32)

C = 5/9 of -72

C = -72 9 x 5 = -40

-40°F = -40°C

Value same in each scale!!

1. Write formula



BO÷ / x+ / -



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(c) C = 5/9(F - 32)

10 = 5/9(F - 32) (x9)

90 = 5(F - 32)

90 = 5F - 160

5F = 90 + 160

5F = 250

F = 50

10°C = 50°F

1. Write formula



BO÷ / x+ / -






EXIT

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Answer

The time, T secs, for a pendulum to swing to & fro is

calculated by the formula

(a) Find T when L = 40m.

(b) Find L when T = 18.84secs.

T = 2 ( )L

10L

where L is the length of the pendulum in metres.

Take = 3.14.


EXIT

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The time, T secs, for a pendulum to swing to & fro is

calculated by the formula


(b) Find L when T = 18.84secs.

T = 2 ( )L

10L

where L is the length of the pendulum in metres.

Take = 3.14.

T = 12.56

L = 90


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Question 3

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(b) Find L when

T = 18.84secs.

T = 2 ( )L

10 T = 2 x 3.14 x ( )4010

T = 2 x 3.14 x 4

T = 12.56

Time is 12.56secs when length is 40m

T = 2 ( )L

10(a)

Take = 3.14.

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Question 3

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(b) Find L when

T = 18.84secs.

T = 2 ( )L

10

Take = 3.14.

T = 2 ( )L

10(b)

2 x 3.14 x = 18.84( )L

10

6.28 x = 18.84( )L

10 ( 6.28)

= 3( )L

10

= 32L10

= 9L10

L = 9 x 10 = 90

Length is 90m when time is 18.84secs

Square bothSides !


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T = 2 x 3.14 x ( )4010

T = 2 x 3.14 x 4

T = 12.56

Time is 12.56secs when length is 40m

T = 2 ( )L

10(a) 1. Write formula




BO÷ / x+ / -


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T = 2 ( )L

10(b)

2 x 3.14 x = 18.84( )L

10

6.28 x = 18.84( )L

10 ( 6.28)

= 3( )L

10

= 32L10

= 9L10

L = 9 x 10 = 90

Length is 90m when time is 18.84secs

Square bothSides !

1. Write formula



BO÷ / x+ / -






EXIT

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Answer

Change the subject of the formula m = 3p2 – k to p.


EXIT

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Change the subject of the formula m = 3p2 – k to p.

p = m + k3


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Question 4

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Change the subject of

the formula

m = 3p2 – k to p.

m = 3p2 – k

3p2 – k = m

3p2 = m + k

p2 = k + m3

Swap sides.

Isolate 3p2

Isolate p2

Isolate p

p = m + k3


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10 = 3x2 – 4

3x2 – 4 = 10

3x2 = 10 + 4

x = 10 + 4 3

Apply the same rules as in a simple equation.

x2 = 10 + 4 3

m = 3p2 – k

3p2 – k = m

3p2 = m + k

p2 = k + m3

Swap sides.

Isolate 3p2

Isolate p2

Isolate p

p = m + k3

EXAMPLE

EXAMPLE

EXAMPLE

EXAMPLE

EXAMPLE


EXIT

Full solution

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Answer

Change the subject of the formula Q = 2kn – 1 to n . 3


EXIT

Full solution

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Change the subject of the formula Q = 2kn – 1 to n . 3

n = 3Q + 3 2k

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Question 5

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Change the subject of the

formula

Q = 2kn – 1 to n . 3

3Q = 2kn - 3

2kn – 3 = 3Q

2kn = 3Q + 3

Q = 2kn – 13

X 3 to eliminate fraction.

Swap sides.

Isolate 2kn .

Isolate n .

n = 3Q + 3 2k

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3 x 10 = 2 x 5 x n - 3

2 x 5 x n – 3 = 3 x 10

2 x 5 x n = (3 x 10) + 3

10 = 2 x 5 x n – 13

n = (3 x 10) + 3 2 x 5

3Q = 2kn - 3

2kn – 3 = 3Q

2kn = 3Q + 3

Q = 2kn – 13

X 3 to eliminate fraction.

Swap sides.

Isolate 2kn .

Isolate n .

n = 3Q + 3 2k

Apply the same rules as in a simple equation.

EXAMPLE

EXAMPLE

EXAMPLE

EXAMPLE

EXAMPLE


UNIT 4 :



EXIT


Logic Diagrams

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2 3 4

Bryan lives in the country and works in town. His journey to work is split into two parts. He can get from home into town by either bus or train. When he arrives in town he can then get to his office by bus or taxi or he can cycle there provided the first part of his journey was made by train.

List all the possible ways he can make his way to work.

LOGIC DIAGRAMS: Question 1

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Answer

Bryan lives in the country and works in town. His journey to work is split into two parts. He can get from home into town by either bus or train. When he arrives in town he can then get to his office by bus or taxi or he can cycle there provided the first part of his journey was made by train.

List all the possible ways he can make his way to work.


EXIT

Full solution

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Combinations are ...

1. Bus-bus 2. Bus-taxi 3. Train-bus

4. Train-taxi 5. Train-cycle

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Question 1

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Bryan’s journey to work is split into two parts. He can get from home into town by either bus or train. He can then get to his office by bus or taxi or he can cycle there provided the first part of his journey was made by train.

bus

train

bus

taxi

bus

taxicycle

bus

train

bus

taxi

bus

taxicycle


1. Bus-bus

2. Bus-taxi

3. Train-bus

4. Train-taxi

5. Train-cycle

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1. Bus-bus

2. Bus-taxi

3. Train-bus

4. Train-taxi

5. Train-cycle

This question may be combined with probability.

If each journey is equally likely what is the probability he travelledby train then bus?

Probability (uses bus) = 35

What is the probability that he Will use a bus at some point in his journey?

Probability (train,bus) = 15


EXIT

MULT INT BY 1.18

STOP

IS THERE TAX EXEMPTION?

START

IS AMOUNT > £10000?

IS AMOUNT > £5000?

INT = 5.3% OF AMOUNT



Yes No

Yes

No

Yes

No

The following flowchart is used to calculate the annual interest on a building society account.

Find the annual interest for a tax-payer with £3850.


EXIT

MULT INT BY 1.18

STOP


START


IS AMOUNT > £5000?




Yes No

Yes

No

Yes

No



= £169.40

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Question 2

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Find the annual interest for

a tax-payer with £3850.

Back to flowchart

Amount is under £5000

so rate = 4.4%

4.4% of £3850 = 0.044 x £3850

= £169.40

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so rate = 4.4%

4.4% of £3850 = 0.044 x £3850

= £169.40

Take care to ensure thatyou follow correct “flow” of

diagram by answering each question carefully.


MULT INT BY 1.18

STOP


START


IS AMOUNT > £5000?




Yes No

Yes

No

Yes

No


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4.4% of £3850 = x £3850

= 0.044 x £3850

4.4100

etc.


so rate = 4.4%

4.4% of £3850 = 0.044 x £3850

= £169.40


EXIT


MULT INT BY 1.28

STOP


START


IS AMOUNT > £5000?




Yes No

Yes

No

Yes

No

Find the annual interest for a non tax-payer with £6700.


EXIT


MULT INT BY 1.28

STOP


START


IS AMOUNT > £5000?




Yes No

Yes

No

Yes

No


= £488.83

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Question 3

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Find the annual interest for

a non tax-payer with £6700.

Back to flowchart

Amount is between £5000 &

£10000 so rate = 5.7%

5.7% of £6700 = 0.057 x £6700

Investor is exempt from tax so int

= 1.28 x £381.90

= £488.83

= £381.90


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£10000 so rate = 5.7%

5.7% of £6700 = 0.057 x £6700


= 1.28 x £381.90

= £488.83

= £381.90

Take care to ensure thatyou follow correct “flow” of

diagram by answering each question carefully.


MULT INT BY 1.28

STOP


START


IS AMOUNT > £5000?




Yes No

Yes

No

Yes

No


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£10000 so rate = 5.7%

5.7% of £6700 = 0.057 x £6700


= 1.28 x £381.90

= £488.83

= £381.90


5.7% of £6700 = x £6700

= 0.057 x £6700

5.7100

etc.


EXIT

The diagram below shows a network of streets near a post office (P). When making deliveries the postman/woman tries to cover all streets without going along the same street more than once if possible.

(a) Explain why you can cover the above network without retracing any part of your route.

(b) List one possible route to illustrate this.

P

GF

E

D

C

B

A


EXIT

The diagram below shows a network of streets near a post office (P). When making deliveries the postman/woman tries to cover all streets without going along the same street more than once if possible.

(a) Explain why you can cover the above network without retracing any part of your route.

(b) List one possible route to illustrate this.

(a)because each vertex/point has an even number of streets meeting at it.

P

GF

E

D

C

B

A

1

23

45

67

89

1011

12

13

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Question 4

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P

GF

E

D

C

B

A

(a)

The network can be traversed without

repeating any route because each

vertex/point has an even number of

streets meeting at it.

(b) One possible route around the network is

P

GF

E

D

C

B

A

1

P G

2

F

3

E

4

D

5

P

6

C

7

D

8

B

9

A

10

C

11

B

12

A

13

P

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P

GF

E

D

C

B

A

1

2

3

4

5

67

89

10

11

12

13

Test for Traversing a network

Network must have no more than two odd vertices.

2 odd1 even traversible

Note: must start on an odd vertex

3

43

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P

GF

E

D

C

B

A

1

2

3

4

5

67

89

10

11

12

13

Test for Traversing a network

Network must have no more than two odd vertices.

NOTtraversible

3

33

2

3

4 odd1 even

End of Unit 4

unit 4 : exit intermediate 2 – additional question bank social arithmetic logic diagrams formulae

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